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3 years ago

Determine the value of x in the figure shown.

Mathematics

1 answer:

garri49 [273]3 years ago

7 0

Hope it will help you..............

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PLEASE HELP ASAP!!!!!!!!!

serg [7]

A.(2,3)
B.(5,5)
C.(7,3)
D.(5,2)

Can someone help meh. Trying to reach Ace tank.

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3 years ago

Henry and Saul live 14 miles apart. Saul and Claire live 20 miles apart

8_murik_8 [283]

So Henry and Saul live 14 miles apart and Clair and Saul live 20 miles apart, that it's really, unless there is more to go off of here

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3 years ago

What is a possible value of sin(theta) when cos2(theta)=0.21

podryga [215]

We know that

cos²(theta)=0.21
sin²(theta)+cos²<span>(theta)=1
</span>sin²(theta)=1-cos²(theta)------> sin²(theta)=1-(0.21)-----> 0.79
sin(theta)=√0.79
sin(theta)=0.89

the answer is
the value of sin(theta) is 0.89

6 0

3 years ago

Assume that the number of messages input to a communication channel in an Exponential distribution with 7 messages arriving in a

NARA [144]

The probability that more than 3 messages will arrive during a 30-second interval is $P(X=n)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^n}{n!}$ .

According to the statement

we have given that the an Exponential distribution with 7 messages arriving in a 10 second period and we have to find the probability that more than 3 messages will arrive during a 30-second interval.

So, For this purpose, we know that the

The probability is the measure of the likelihood of an event to happen. It measures the certainty of the event.

And the given information is that :

3 messages will arrive during a 30-second interval.

Then

Probability = P(X=1) + P(X=2) + P(X=3).Then

The probability become according to the exponential distribution:

$P(X=1)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^1}{1!}\\P(X=2)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^2}{2!}\\P(X=3)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^3}{3!}\\$

And then substitute the values in it then

$Probability = \dfrac{e^{-0.3 \times 20} (0.3 \times 20)^1}{1!}\ +\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^2}{2!}\ + \dfrac{e^{-0.3 \times 20} (0.3 \times 20)^3}{3!}\\$

This is the probability.

So, The probability that more than 3 messages will arrive during a 30-second interval is $P(X=n)=\dfrac{e^{-0.3 \times 20} (0.3 \times 20)^n}{n!}$ .

Learn more about probability here

brainly.com/question/24756209

#SPJ4

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2 years ago

The beckelys purchased 17 gallons of gas on their trip. The gas cost 4.72 per gallons what was the total cost ?

valentina_108 [34]

I'm not entirely Sure but I Believe the answer is

$80.24

6 0

3 years ago